# Why is it so hard to find a closed form for $\sum_{n=1}^\infty \frac{1}{n^3}$?

The series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6} \text{ and } \sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$$

I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$\sum_{n=1}^\infty \frac{1}{n^3}=\zeta(3)\approx 1.2020569...$$

I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.

For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.

• Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it. Commented Dec 30, 2016 at 3:52
• The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form. Commented Dec 30, 2016 at 3:52
• That one can solve the even cases can be traced back to the fact that when you plug $\pi$ into the Taylor expansion of $\sin(x)$ you should get 0. There is nothing similar for the odd cases. Commented Dec 30, 2016 at 4:00
– user371838
Commented Dec 30, 2016 at 4:11
• See this similar question we can compute a closed-form of $\zeta(2n)$ because it is the derivative of $\cos(z)/\sin(z)$ at $z=1/2$, but $\zeta(2n+1)$ is the derivative of $\Gamma'(z)/\Gamma(z)$ at $z=1/2$ which is much more complicated Commented Dec 30, 2016 at 10:29

Roger Apéry proved in 1978 that $$\zeta(3)=\displaystyle \sum_{n=1}^{\infty} n^{-3}$$ is irrational, however the irrationality of $$\displaystyle\frac{\zeta(3)}{\pi^3}$$ is still an open problem. No closed form for its value is currently known.

There are however some formulas expressing $$\zeta(3)$$ (and other odd zeta values) in terms of powers of $$\pi$$, but these are not closed forms. The most well known ones are due to Plouffe and Borwein & Bradley:

\begin{aligned} \zeta(3)&=\frac{7\pi^3}{180}-2\sum_{n=1}^\infty \frac{1}{n^3(e^{2\pi n}-1)},\\ \sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} &= -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right). \end{aligned}

Moreover, in this Math.SE post we have:

$$\frac{3}{2}\,\zeta(3) = \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}.$$

You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.

It can be shown that $$\frac{\sin \pi x}{\pi x}=\prod_{k\ge 1}(1-\frac{x^2}{k^2})$$. The left-hand side's Maclaurin series can be used to write $$\pi^{2n}$$ in terms of $$\zeta(2k),\,1\le k\le n$$. There is no similar theoretical treatment of $$\zeta(m)$$ for odd $$m\ge 3$$.